Question

(5x^4-x^3+4x^2+8)/(x^3-8) интеграл

Answer 1

$\int \frac{5x^4-x^3+4x^2+8}{x^3-8}=\int (5x-1+\frac{4x^2+40x}{(x-2)(x^2+2x+4)})dx=\\\\=\frac{5x^2}{2}-x+\int \frac{2\, dx}{x-2}+\int \frac{-4x+16}{x^2+2x+4}dx=2,5x^2-x+2ln|x-2|-\\\\-4\int \frac{(x-4)dx}{(x+1)^2+3}=2,5x^2-x+2ln|x-2|-4\int \frac{(t-5)dt}{t^2+3}=[t=x+1]=\\\\=2,5x^2-x+2ln|x-2|-2\int \frac{2t\, dt}{t^2+3}+20\int \frac{dt}{t^2+3}=[2t\, dt=d(t^2+3)]=\\\\=2,5x^2-x+2ln|x-2|-2ln|t^2+3|+\frac{20}{\sqrt3}\cdot arctg\frac{t}{\sqrt3}+C=\\\\=2,5x^2-x+2ln|x-2|-2ln|x^2+2x+4|+\frac{20}{\sqrt3}\cdot arctg\frac{x+1}{\sqrt3}+C$

$P.S.\; \; \frac{4x^2+40x}{(x-2)(x^2+2x+4)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+2x+4}\\\\4x^2+40x=Ax^2+2Ax+4A+Bx^2-2Bx+Cx-2C\\\\x=2:\; \; A=\frac{96}{12}=8\\\\x^2|\; 4=A+B\; ,\; \; B=4-A=-4\\x^0|\; 0=4A-2C\; ,\; \; C=2A=16\\\\\frac{4x^2+40x}{(x-2)(x^2+2x+4)}=\frac{8}{x-2}+\frac{-4x+16}{x^2+2x+4}$